Problem 70 - A Complete Resource Book in Mathematics for JEE Main

Question

If the equation of the locus of a point equidistant from the points \(\left(a_{1}, b_{1}\right)\) and \(\left(a_{2}, b_{2}\right)\) is \(\left(a_{1}-a_{2}\right) x+\left(b_{1}-b_{2}\right) y\) \(+c=0\), then the value of \(^{\circ} c^{\prime}\) is \(\quad\) (A) \(\frac{1}{2}\left(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}\right)\) (B) \(a_{1}^{2}+a_{2}^{2}-b_{1}^{2}-b_{2}^{2}\) (C) \(\frac{1}{2}\left(a_{1}^{2}+a_{2}^{2}-b_{1}^{2}-b_{2}^{2}\right)\) (D) \(\sqrt{a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}}\)

Step-by-Step Solution

Answer

The correct option is (A) \(\frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)\).

Step 1: Understanding the Locus Equation

The equation given is \[(a_1 - a_2) x + (b_1 - b_2) y + c = 0\]This represents the locus of points equidistant from two given points \((a_1, b_1)\) and \((a_2, b_2)\). The line representing the locus will be the perpendicular bisector of the segment joining these points.

Step 2: Deriving the Perpendicular Bisector Equation

The perpendicular bisector of the segment connecting \((a_1, b_1)\) and \((a_2, b_2)\) is derived by setting the distance from any point \((x, y)\) to \((a_1, b_1)\) equal to its distance to \((a_2, b_2)\). This leads to the equation:\[\sqrt{(x - a_1)^2 + (y - b_1)^2} = \sqrt{(x - a_2)^2 + (y - b_2)^2}\]

Step 3: Squaring Both Sides and Expanding

Square both sides of the derived equation from Step 2 and expand:\[(x - a_1)^2 + (y - b_1)^2 = (x - a_2)^2 + (y - b_2)^2\]This simplifies to:\[x^2 - 2a_1x + a_1^2 + y^2 - 2b_1y + b_1^2 = x^2 - 2a_2x + a_2^2 + y^2 - 2b_2y + b_2^2\]

Step 4: Simplifying the Resultant Equation

Cancel out common terms on both sides (i.e., \(x^2\) and \(y^2\)) and rearrange the terms:\[-2a_1 x - 2b_1 y + a_1^2 + b_1^2 = -2a_2 x - 2b_2 y + a_2^2 + b_2^2\]

Step 5: Arranging the Linear Equation

Rearrange the equation from Step 4 to collect term coefficients:\[2(a_2 - a_1)x + 2(b_2 - b_1)y = a_2^2 + b_2^2 - a_1^2 - b_1^2\]

Step 6: Identifying the Constant Term

The equation from Step 5 simplifies to match the form given in the problem statement:\[(a_1 - a_2)x + (b_1 - b_2)y = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)\]This indicates that the constant term \(c\) is:\[c = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)\]

Step 7: Matching the Correct Option

Compare the derived expression for \(c\) with the given options:*Option (A) is \(\frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)\), which matches our derived expression.*

Key Concepts

Perpendicular Bisector

A perpendicular bisector of a line segment is a line that divides the segment into two equal parts at a right angle. It's a key concept when discussing the locus of points equidistant from two fixed points. Consider two points, \((a_1, b_1)\) and \((a_2, b_2)\). The line that is equidistant from both these points is their perpendicular bisector. This bisector can be found by ensuring that any point on it is exactly the same distance from both points. This leads us to an important method in geometry: by solving for the perpendicular bisector, we find the equation representing all possible equidistant points, effectively identifying the locus.

• The midpoint of the segment connecting \((a_1, b_1)\) and \((a_2, b_2)\) is \(\left(\frac{a_1+a_2}{2}, \frac{b_1+b_2}{2}\right)\).
• The slope of the line passing through these two points is \(\frac{b_2-b_1}{a_2-a_1}\).
• The perpendicular bisector will have a slope that is the negative reciprocal of the original slope, \(-\frac{a_2-a_1}{b_2-b_1}\), if the line isn't vertical.

By knowing the slopes, we can derive the equation of the perpendicular bisector, providing the foundation for the locus equation.

Distance Formula

The distance formula is a useful tool in coordinate geometry for determining how far apart two points are. This is essential when working with perpendicular bisectors as it helps establish equidistance. To find the perpendicular bisector, we use the fact that any point on it is equidistant from the two given points. For points \((x_1, y_1)\) and \((x_2, y_2)\), the distance is computed by:

• Using the formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

When deriving the equation for a perpendicular bisector, you ensure distances from an arbitrary point \((x, y)\) on the bisector to the two given points are equal:

• \(\sqrt{(x - a_1)^2 + (y - b_1)^2} = \sqrt{(x - a_2)^2 + (y - b_2)^2}\)

By simplifying this equation, you derive the linear equation that describes the location of the points on the perpendicular bisector.

Coordinate Geometry

Coordinate geometry allows us to analyze geometric situations algebraically. It assigns numerical coordinates to points on a plane, simplifying the study of geometrical problems such as finding the locus of points equidistant from two specific points. By translating geometric problems into algebra, you can solve for unknowns more easily, such as finding the equation of a perpendicular bisector.To solve for the locus of a point equidistant from \((a_1, b_1)\) and \((a_2, b_2)\), we derived it from coordinate geometry principles:

• Apply the distance formula to enforce equidistance, forming a system of equations.
• Through algebraic manipulations (squaring both sides, canceling terms), establish the linear equation \((a_1 - a_2)x + (b_1 - b_2)y + c = 0\).
• The constant \(c\) is often derived by collecting terms, as shown in the given problem: \(c = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)\).

This approach illustrates how geometric concepts can seamlessly translate into coordinate geometry, revealing values such as \(c\) and offering insightful solutions.